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Alex787 [66]
3 years ago
15

Neil studied how a pulley helps move an object. Neil tied a rope to an object with a mass of 10 kilograms (kg) and attached it t

o a pulley. He then pulled down on the rope. When Neil pulled down on the rope, which did he most likely learn about the force required to lift the object? Group of answer choices
1. It was half of the weight of the object.
2. It was equal to the weight of the object
3. It was two times the weight of the object.
4. It was three times the weight of the object.
Physics
1 answer:
SashulF [63]3 years ago
3 0

Answer:2

Explanation:

Given

mass of object m=10\ kg

weight of object =mg

Weight=10\times 9.8

If body is being pulled with constant velocity then the net force on body is zero i.e.

Force=weight of object

F=98\ kg

force is equal to weight of object

option 2 is correct

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9. Consider the elbow to be flexed at 90 degrees with the forearm parallel to the ground and the upper arm perpendicular to the
mojhsa [17]

Answer:

Moment about SHOULDER  ∑ τ = 3.17 N / m,

Moment respect to ELBOW   Στ= 2.80 N m

Explanation:

For this exercise we can use Newton's second law relationships for rotational motion

         ∑ τ = I α

   

The moment is requested on the elbow and shoulder at the initial instant, just when the movement begins.

They indicate the angular acceleration, for which we must look for the moments of inertia of the elements involved

The mass of the forearm with the included weight is approximately 2.3 kg, with a length of about 50cm

Moment about SHOULDER

          ∑ τ = I α

           I = I_forearm + I_sphere

the forearm can be approximated as a fixed bar at one end

            I_forearm = ⅓ m L²

the moment of inertia of the mass in the hand, let's approach as punctual

            I_mass = m L²

we substitute

           ∑ τ = (⅓ m L² + M L²) α

let's calculate

          ∑ τ = (⅓ 2.3 0.5² + 0.5 0.5²) 10

           ∑ τ = 3.17 N / m

Moment with respect to ELBOW

In this case, the arm exerts an upward force (muscle) that is about 3 cm from the elbow

         Στ = I α

         I = I_ forearm + I_mass

         I = ⅓ m (L-0.03)² + M (L-0.03)²

         

let's calculate

        i = ⅓ 2.3 0.47² + 0.5 0.47²

        I = 0.2798 Kg m²

        Στ = 0.2798 10

        Στ= 2.80 N m

3 0
3 years ago
The atoms of a molecule come from two or more?
AlladinOne [14]
Compounds. two or more compounds.
7 0
3 years ago
A swimmer, capable of swimming at a speed of 1.0 m/s in still water (i.e., the swimmer can swim with a speed of 1.0 m/s relative
Vesnalui [34]
If the current takes him downstream we must find the resultant vector of the velocities: V res= \sqrt{1^{2}+0.91^{2}  } = \sqrt{1.8281}= 1.3520747 Then if the river is 3000 m-wide the swimmer will have to pass:
 1.3520747 · 300 = 4056.14 m                t = 4056.14 m : 1 m/s
a ) It takes 4056.15 seconds ( 1 hour 7 minutes and 36 seconds ) to cross the river.  
b ) 0.91 · 3000 = 2730 m
He will be 2730 m downstream.
5 0
3 years ago
What type of lens does a camera contain?
Savatey [412]
A camera lens (also known as photographic lens or photographic objective) is an optical lens or assembly of lenses used in conjunction with a camera body and mechanism to make images of objects either on photographic film or on other media capable of storing an image chemically or electronically.
7 0
4 years ago
Read 2 more answers
Q = cmAT
castortr0y [4]

Answer: Q=3000 cal

Explanation:

We are given the following formula:

Q=m. c. \Delta T   (1)

Where:

Q=3000 cal is the amount of heat

m=300g  is the mass  of water

c=1 cal/g \°C  is the specific heat of water

\Delta T  is the variation in temperature, which in this case is  \Delta T=30\°C-20\°C=10\°C  

Rewriting equation (1) with the known values at the right side, we will prove the result is 3000 cal:

Q=(300g)(1 cal/g \°C)(10\°C)   (2)

Q=3000 cal   This is the result

8 0
3 years ago
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